Solve by factoring

I'm not even sure if re-writing it that way helps. I was thinking I could also do this:

(x^2)^2 + x - 18 = 0

Replace (x^2) with y:

y^2 + y^(1/2) - 18 = 0

I still don't know if that would even help. I plugged the problem into an equation solving software and it said the only answer was 2 (unfortunately it doesn't explain how to solve it). By the way, if I use Descartes Rule of Signs, it says there should be a negative root (-x)^4 - x - 18. There is 1 sign change, but the software made no mention of another root. Did I use the rule of signs technique wrong?

I'm not even sure if re-writing it that way helps. I was thinking I could also do this:

(x^2)^2 + x - 18 = 0

Replace (x^2) with y:

y^2 + y^(1/2) - 18 = 0

I still don't know if that would even help. I plugged the problem into an equation solving software and it said the only answer was 2 (unfortunately it doesn't explain how to solve it). By the way, if I use Descartes Rule of Signs, it says there should be a negative root (-x)^4 - x - 18. There is 1 sign change, but the software made no mention of another root. Did I use the rule of signs technique wrong?

Since x = 2 is easily noted to be a zero of your quartic equation, it can be factorised as .

The cubic has only one real zero, and it will not be an easy task to get that zero in exact form (you might need to use the Cardano method).

Since x = 2 is easily noted to be a zero of your quartic equation, it can be factorised as .

The cubic has only one real zero, and it will not be an easy task to get that zero in exact form (you might need to use the Cardano method).

Is there any method I can use besides the rational root theorem that can tell me that one of the roots is 2? I only knew one of the roots was 2 from using a calculator. And was I right in my guess that there is only 1 positive root and 1 negative root?